A certain transmission line is constructed from two thin metal "rib-bons," of width w, a very small distance$h\ll w$ apart. The current travels down one strip and back along the other. In each case, it spreads out uniformly over the surface of the ribbon.

(a) Find the capacitance per unit length, C .

(b) Find the inductance per unit length, L .

(c) What is the product LC , numerically?[ L and C will, of course, vary from one kind of transmission line to another, but their product is a universal constantcheck, for example, the cable in Ex. 7.13-provided the space between the conductors is a vacuum. In the theory of transmission lines, this product is related to the speed with which a pulse propagates down the line: $\mathbf{v}\mathbf{=}\mathbf{1}\mathbf{/}\sqrt{\mathbf{LC}}$.]

(d) If the strips are insulated from one another by a non-conducting material of permittivity $\mathbf{\epsilon }$and permeability $\mathbf{\epsilon }$and permeability $\mathbf{\mu }$, what then is the product LC ? What is the propagation speed? [Hint: see Ex. 4.6; by what factor does L change when an inductor is immersed in linear material of permeability$\mathbf{\mu }$?]

The wis the width of ribbons.

The h is the separation between two ribbons.

The I is the length of the ribbon.

Write the formula ofcapacitance per unit length.

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$ …… (1)

Here, C is the capacitance and V is the voltage.

Write the formula ofinductance per unit length.

$\mathit{\varphi }\mathbf{=}\mathbf{LI}$ …… (2)

Here, L is inductance per unit length and I is the length of the ribbon.

Write the formula ofpropagation speed.

$\mathbf{v}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{\epsilon }}_{\mathbf{0}}}}$ …… (3)

Here,${\mathit{\epsilon }}_{\mathbf{0}}$ is permittivity and${\mathit{\mu }}_{\mathbf{0}}$ is permeability.

Write the formula ofpropagation speed.

$\mathbf{L}\mathbf{=}\frac{\mathbf{\mu h}}{\mathbf{w}}$ …… (4)

Here,$\mathit{\mu }$ is permeability, h is the separation between two ribbons and w is the width of ribbons.

Now we discuss for a parallel plate capacitor.

The electric field between the plates of the capacitor is

$E=\frac{\sigma }{{\epsilon }_0}$

The potential difference between the plates of capacitor is

$V=Eh$

Then $V=\left(\frac{\sigma }{{\epsilon }_0}\right)h$

But role="math" localid="1657533956022" $\sigma =\mathrm{Charge}\mathrm{density}=\frac{Q}{wI}$

Then $V=\frac{1}{{\epsilon }_0}\left(\frac{Q}{wI}\right)h$

Determine the capacitance per unit length.

Substitute$\frac{1}{{\epsilon }_0}\left(\frac{Q}{wI}\right)h$ for v into equation (1).

$$\begin{gathered} C=\frac{Q}{{\displaystyle \frac{1}{{\epsilon }_0}}\left({\displaystyle \frac{Q}{wI}}\right)h} \\ =\frac{{\epsilon }_{0}wI}{h} \end{gathered}$$

Therefore, the value of capacitance per unit length is$\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{w}\mathrm{I}}{\mathrm{h}}$ .

We know that magnetic field.

$$\begin{gathered} B={\mu }_{0}k \\ k=\frac{I}{w} \end{gathered}$$

Then $$\begin{gathered} B=\frac{{\mu }_{0}I}{w} \\ \varphi =BhI \end{gathered}$$

Then $=\frac{{\mu }_{0}I}{w}hL$

Determine inductance per unit length.

Substitute$\frac{{\mu }_{0}IhI}{w}$ for$\varphi$ into equation (2).

$$\begin{gathered} \frac{{\mu }_{0}IhI}{w}=LI \\ L=\frac{{\mu }_{0}IhI}{w} \end{gathered}$$

Then inductance per unit length.

$$\begin{gathered} \frac{L}{I}=L \\ =\frac{{\mu }_{0}h}{w} \end{gathered}$$

Therefore, the value of inductance per unit length is $L=\frac{{\mu }_{0}h}{w}$.

Derive the propagation speed as follows:

$$\begin{gathered} LC=\frac{{\epsilon }_{0}w}{h}\times \frac{{\mu }_{0}h}{w} \\ ={\mu }_0{\epsilon }_0 \\ =\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right) \\ =1.112\times {10}^{-17}{\mathrm{s}}^2/{\mathrm{m}}^2 \end{gathered}$$

Determine the propagation speed.

Substitute $\mathrm{LC}$for$\sqrt{{\mu }_0{\epsilon }_0}$ into equation (3).

$$\begin{gathered} v=\frac{1}{\sqrt{\mathrm{LC}}} \\ =2.999\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the value of propagation speed $v=2.99\times {10}^8\mathrm{m}/\mathrm{s}$.

As a non-conducting substance with permittivity $\epsilon$and permeability $\mu$separates the strips from one another as follows:

$$\begin{gathered} D=\sigma \\ E=\frac{D}{\epsilon } \\ E=\frac{\sigma }{\epsilon } \end{gathered}$$

Then solve as:

$$\begin{gathered} C=\frac{\epsilon w}{h} \\ H=K \\ B=\mu H \\ B=\mu K \end{gathered}$$

Determine propagation speed.

Substitute$\epsilon$for$\frac{h}{w}$into equation (4).

localid="1657535319966" $$\begin{gathered} LC=\epsilon \mu \\ v=\frac{1}{\sqrt{\epsilon \mu }} \end{gathered}$$

Therefore, the value of propagation speed $v=\frac{1}{\sqrt{\epsilon \mu }}$.